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2 years ago

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A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

istance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?

1 answer:

zvonat [6]2 years ago

5 0

Answer:

<u /> $D_l=d$ <u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

<u /> $D_l=d$ <u />

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Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

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$KE _A$ is the kinetic energy at A which is mathematically represented as

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